Question

A coil of area \( 12 \text{ cm}^2 \) has \( 250 \) turns. Magnetic field of \( 0.2 \text{ Wb/m}^2 \) is perpendicular to the plane of the coil. The field is reduced to \( 0.1 \text{ Wb/m}^2 \) in \( 0.1 \text{ second} \). The magnitude of induced e.m.f. in the coil is

• A. 0.2 V
• B. 0.3 V
• C. 3 V
• D. 6 V

Hint:

- Always convert area to $\text{m}^2$ - $\Phi = BA$ when field is perpendicular

Answer 1

The Correct Option is B

Concept:
Faraday’s law of electromagnetic induction: \[ \mathcal{E} = N \frac{\Delta \Phi}{\Delta t} \quad \text{where} \quad \Phi = BA \]

Step 1: Convert area into SI units.
\[ A = 12\ \text{cm}^2 = 12 \times 10^{-4} = 1.2 \times 10^{-3}\ \text{m}^2 \]

Step 2: Find change in magnetic field.
\[ \Delta B = 0.2 - 0.1 = 0.1\ \text{Wb/m}^2 \]

Step 3: Calculate change in flux.
\[ \Delta \Phi = A \cdot \Delta B = (1.2 \times 10^{-3})(0.1) = 1.2 \times 10^{-4}\ \text{Wb} \]

Step 4: Apply Faraday’s law.
\[ \mathcal{E} = \frac{N \Delta \Phi}{\Delta t} = \frac{250 \times 1.2 \times 10^{-4}}{0.1} \] \[ \mathcal{E} = \frac{3 \times 10^{-2}}{0.1} = 0.3\ \text{V} \]